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I am trying to prove the Nested Interval Theorem, which is:

Given nested closed intervals of Real Numbers $$[a_1,b_1]\supset[a_2,b_2]\supset\cdots\supset[a_n,b_n]\supset\cdots$$ and $\lim_{n\to\infty}|a_n-b_n|=0$, there exists an element $x_0\in \bigcap_{i=1}^\infty [a_i,b_i]$ and $x_0$ is unique.

I think my proof is wrong somewhere. Could anybody tell me where the mistake is?

Proof: (By contradiction.) Suppose $\cap[a_n,b_n]=\varnothing$, then $\exists n_1,n_2$ ($n_2>n_1$), such that $[a_{n_1},b_{n_1}]\cap[a_{n_2},b_{n_2}]=\varnothing$, but $[a_{n_1},b_{n_1}]\supset [a_{n_2},b_{n_2}]$ and $b_{n_2}-a_{n_2}\neq0$, contradiction.

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  • $\begingroup$ I think for the statement you're proving, that you want to add that $a_n \rightarrow c$ and $b_n \rightarrow c$ for some real number $c$? $\endgroup$ – benguin Oct 7 '17 at 17:07
  • $\begingroup$ There is another statement: $lim_{n->\infty}|b_n-a_n|=0$. sorry for forgetting it. $\endgroup$ – user2874626 Oct 8 '17 at 3:07
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The statement fails to be true. Take $I_n=[-1-1/n, 1+1/n]$ for $n\in \mathbf{N}_{\ge 1}$. $\bigcap_{n=1}^\infty I_n=[-1,1].$ This set contains infinitely many elements.

Your proof does not address the statement, because you are trying to show existence of an element in the intersection without uniqueness.

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Given your setup $[a_1,b_1]\supset[a_2,b_2]\supset ...\supset[a_n,b_n]\supset...\Rightarrow \exists x_0\in \cap[a_n,b_n]$ and $x_0$ is unique.

Your conclusion is too strong. Take, for example, $a_n=2,$ and $b_n=3+\frac{1}{n}.$ Then $$\bigcap_{i=1}^\infty ~[a_i,b_i]=[2,3].$$

Plianly, there is no unique $x\in[2,3].$

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